Question

The perimeter of a sector is constant. If its area is to be maximum, then the sectorial angle is
A. \[\dfrac{\pi }{6}\]
B. \[\dfrac{\pi }{4}\]
C. \[{4^C}\]
D. \[{2^C}\]

Answer 1

Hint: We will first let the radius be \[r\] and sectorial angle be \[\theta \]. As the perimeter of a sector is constant so, we will find the perimeter of a sector and assume it equal to the constant. From here we will evaluate the value of radius and then we will substitute the value of radius in the formula of area of the sector. Next, we will differentiate it both sides with respect to \[\theta \]. As the area needs to be maximum so, we will put the differentiate value equal to zero and obtain the value of \[\theta \]. Next, we will again differentiate the area and substitute the obtained value of \[\theta \]. If the differentiated value is negative, then the area is maximum and thus the value of \[\theta \] is correct otherwise not.

Complete step by step Answer:

We will first let that the radius of the circle is \[r\] and the sectorial angle be \[\theta \].
Consider the figure,

As we know that the perimeter of the sector is given by \[2r + r\theta \] and according to the question, the perimeter of the sector is constant.
So, we get,
\[2r + r\theta = k\]
From the above expression, we will evaluate the value of \[r\],
\[ \Rightarrow r = \dfrac{k}{{2 + \theta }}\]
Now, we will suppose that the area of the sector is denoted by \[A\] and area of the sector is given by \[A = \dfrac{1}{2}{r^2}\theta \]
Here, we will substitute the value of radius obtained above in the formula of area,
We get,
\[ \Rightarrow A = \dfrac{{{k^2}}}{2} \cdot \dfrac{\theta }{{{{\left( {\theta + 2} \right)}^2}}}\]
Now, we will differentiate the area on both sides with respect to \[\theta \] using the formula, \[\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}\],
Thus, we get,
\[
   \Rightarrow \dfrac{{dA}}{{d\theta }} = \dfrac{{{k^2}}}{2}\left\{ {\dfrac{{{{\left( {\theta + 2} \right)}^2} - 2\theta \left( {\theta + 2} \right)}}{{{{\left( {\theta + 2} \right)}^4}}}} \right\} \\
   \Rightarrow \dfrac{{dA}}{{d\theta }} = \dfrac{{{k^2}}}{2}\left( {\dfrac{{\left( {\theta + 2} \right)\left( {\theta + 2 - 2\theta } \right)}}{{{{\left( {\theta + 2} \right)}^4}}}} \right) \\
   \Rightarrow \dfrac{{dA}}{{d\theta }} = \dfrac{{{k^2}}}{2}\left( {\dfrac{{2 - \theta }}{{{{\left( {\theta + 2} \right)}^3}}}} \right) - - - - - {\text{eq}}\left( 1 \right) \\
 \]
As it is given in the question, that the area has to be maximum so, we will put the value of \[\dfrac{{dA}}{{d\theta }}\] written as equation (1) equal to zero.
Thus, we get,
\[
   \Rightarrow \dfrac{{dA}}{{d\theta }} = 0 \\
   \Rightarrow \dfrac{{{k^2}}}{2}\left( {\dfrac{{2 - \theta }}{{{{\left( {\theta + 2} \right)}^3}}}} \right) = 0 \\
   \Rightarrow 2 - \theta = 0 \\
   \Rightarrow \theta = 2 \\
 \]
Next, we will again differentiate the equation (1) with respect to \[\theta \], we get,
\[
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} = \dfrac{{{k^2}}}{2}\left[ {\dfrac{{2\left( { - 3} \right)}}{{{{\left( {\theta - 2} \right)}^4}}} - \dfrac{{{{\left( {\theta + 2} \right)}^3} \times 1 - \theta \times 3{{\left( {\theta + 2} \right)}^2}}}{{{{\left[ {{{\left( {\theta + 2} \right)}^3}} \right]}^2}}}} \right] \\
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} = \dfrac{{{k^2}}}{2}\left[ {\dfrac{{ - 6}}{{{{\left( {\theta + 2} \right)}^4}}} - \dfrac{{\theta + 2 - 3\theta }}{{{{\left( {\theta + 2} \right)}^4}}}} \right] \\
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} = \dfrac{{ - {k^2}}}{2}\left[ {\dfrac{6}{{{{\left( {\theta + 2} \right)}^4}}} + \dfrac{{2 - \theta }}{{{{\left( {\theta + 2} \right)}^4}}}} \right] \\
 \]
Now, we will substitute the obtained value of \[\theta = 2\] in the above expression,
Thus, we have,
\[
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} = \dfrac{{ - {k^2}}}{2}\left[ {\dfrac{6}{{{{\left( {2 + 2} \right)}^4}}} + \dfrac{{2 - 2}}{{{{\left( {2 + 2} \right)}^4}}}} \right] \\
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} = \dfrac{{ - {k^2}}}{2}\left[ {\dfrac{6}{{{{\left( 4 \right)}^4}}} + 0} \right] \\
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} = \dfrac{{ - 3{k^2}}}{{256}} \\
   \Rightarrow \dfrac{{{d^2}A}}{{d{\theta ^2}}} < 0 \\
 \]
From this, we can conclude that the area is maximum as the double derivative of the area is negative at \[\theta = {2^C}\].
Hence, option D is correct.

Note: We have to remember that the area is maximum when the double derivative is negative and minimum when the double derivative is positive. We need to find the value of \[\theta \] to determine that the double derivative is negative or positive. We have to remember the formula of area, \[A = \dfrac{1}{2}{r^2}\theta \] and the perimeter of the sector is given by \[2r + r\theta \].